Is there an object such that its unique existence follows from choice axiom, while its existence cannot be proven without choice axiom? Let $\varphi$ be a formula, and suppose $\text{ZFC}\vdash \exists x\varphi(x)\wedge\forall x\forall y(\varphi(x)\wedge\varphi(y)\to x=y)$
Is $\text{ZF}\vdash \exists x\varphi(x)$ true?
 A: I posted this originally as a comment to Alex's answer but, at his suggestion, I am expanding it into a proper answer.

This situation actually occurs in practice in infinitary combinatorics: we use the axiom of choice to establish the existence of an object, but its uniqueness then follows without further appeals to choice. I point this out to emphasize that this is actually a common and interesting phenomenon, rather than the result of metamathematical tricks, as one might erroneously conclude from the examples in the other answers.
For a simple example, from choice one can prove that there are uncountable regular cardinals. Once we know that they exist, it is obvious that the least such cardinal is unique. However, $\mathsf{ZF}$ is not enough to prove its existence. This example is perhaps a bit misleading in that, under $\mathsf{ZFC}$, $\aleph_1$ is the least uncountable regular cardinal, but $\mathsf{ZF}$ suffices to prove the existence of $\aleph_1$. What it cannot prove is its regularity.
More interesting examples identify objects whose existence requires the axiom of choice and have no counterpart in the choiceless setting in general. This was studied in some detail, in the context of pcf theory, in

MR2078366 (2005k:03105). Cummings, James; Foreman, Matthew; Magidor, Menachem. Canonical structure in the universe of set theory. I. Ann. Pure Appl. Logic 129 (2004), no. 1-3, 211–243.

From the introduction: "Of particular interest are invariants which are canonical, in the sense that the Axiom of Choice is needed to show that they exist, but once shown to exist they are independent of the choices made. For example the uncountable regular cardinals are canonical in this sense."
The examples studied in the paper have to do with the combinatorics at the successor of a singular cardinal $\mu$. The simplest kind of objects considered in this setting were introduced by Shelah: these are certain nice collections of points in $\mu^+$, that he dubbed good and approachable. These collections are unique modulo the nonstationary ideal.
A: Thankyou for your comment @Mauro ALLEGRANZA
I totally got your point.
Let $\varphi$ be a formula such that $\text{ZF}\vdash \exists_{unique} x\varphi(x)$
Then $\text{ZF}\not\vdash\exists x(\varphi(x)\wedge\text{AC})$ while $\text{ZFC}\vdash \exists_{unique} x(\varphi(x)\wedge \text{AC})$
A: Let $\varphi(x)$ say: (AC and $x=\varnothing$) or ($\lnot$AC and $x\neq x$).
ZFC proves the existence of a unique $x$ satisfying $\varphi(x)$, namely $\varnothing$.
ZF does not prove the existence of an $x$ satifying $\varphi(x)$. If it did, it would prove (AC and $x=\varnothing$) and hence prove AC, but AC is independent of ZF.
Edit: Oops, the second disjunct is not necessary here. (AC and $x=\varnothing$) works fine, as shown in Yuz's answer.
A: While these formal statements are made, here are some real life examples,
"The cardinality of a Hamel basis of $\Bbb R$ over $\Bbb Q$", is a unique and well-defined object once some Hamel basis exists (we can prove that a Hamel basis must always have the cardinality of the continuum). Nevertheless, it is consistent with $\sf ZF$ that no such basis exists.
Similarly we can talk about "the cardinal of a free ultrafilter on $\omega$".
In general, any type of object whose cardinality in $\sf ZFC$ is unique (indeed the above cases are even better: the uniqueness of cardinality follows from $\sf ZF+$"The object exists") but $\sf AC$ is necessary for the existence of the objects, is a good target.
